Posted:

**Fri Oct 22, 2004 8:55 pm**Meh, I distinctly remember doing that problem in Algebra II class in high school lol. Doesn't mean I'm going to bother working it out again now though. :p

Page **5** of **8**

Posted: **Fri Oct 22, 2004 8:55 pm**

Meh, I distinctly remember doing that problem in Algebra II class in high school lol. Doesn't mean I'm going to bother working it out again now though. :p

Posted: **Fri Oct 22, 2004 9:06 pm**

long long ago .....

i don't know if it's right but :

obvious the solution is not 25mph

say the car runs for an hour with 30mph in one direction and than with 20mph in the other (which needs 1.5hours) so

1h * 30mph + 1.5*20mph = 60miles,

the car has gone 60miles in 2.5 hours, so the average speed was 24mph.

plz, don't laugh, it's really long ago

i don't know if it's right but :

obvious the solution is not 25mph

say the car runs for an hour with 30mph in one direction and than with 20mph in the other (which needs 1.5hours) so

1h * 30mph + 1.5*20mph = 60miles,

the car has gone 60miles in 2.5 hours, so the average speed was 24mph.

plz, don't laugh, it's really long ago

Posted: **Fri Oct 22, 2004 11:53 pm**

24 mph

Edit: oops. Next time i should make sure I read the entire thread...

So, here's my solution.

Let d be the (unspecified/unimportant) distance

v1 the velocity for one route (30 mph), v2 the route back (20 mph)

So it takes t = D/v1 + D/v2 time in total for the whole trip. The average speed therefor is v = 2D/t = 2D/(D/v1 + D/v2).

Multiply numerator and denominator with 1/D => v = 2/(1/v1 + 1/v2)

Edit: oops. Next time i should make sure I read the entire thread...

So, here's my solution.

Let d be the (unspecified/unimportant) distance

v1 the velocity for one route (30 mph), v2 the route back (20 mph)

So it takes t = D/v1 + D/v2 time in total for the whole trip. The average speed therefor is v = 2D/t = 2D/(D/v1 + D/v2).

Multiply numerator and denominator with 1/D => v = 2/(1/v1 + 1/v2)

Posted: **Tue Oct 26, 2004 4:18 pm**

Well i figured out the answer wasn't 25. But like hell i'm going to do the maths to work out what the proper answer is. hehe. leave that to the other suckers around here... hehe

another one:

Forward I'm heavy,

Backward I'm not.

What am I?

another one:

Forward I'm heavy,

Backward I'm not.

What am I?

Posted: **Tue Oct 26, 2004 4:24 pm**

a ton (1000 kg)

Posted: **Tue Oct 26, 2004 6:44 pm**

Ghetto rig all three lamps to turn on with only one of the switches. Problem solved.

-----

Here is one of my favorite 'riddles' :

Say that a certain game show has a feature where a contestant is presented with three doors. A worthwhile prize is behind only one of the doors. The contestant is instructed to pick one of the three doors. Afterwards, the host always opens one of the remaining doors that does not have the prize behind it, and then questions the contestant about whether they want to remain with their chosen door or switch to the remaining door.

The question - What is the rule of thumb that provides a higher 'statistical' chance of winning?

sorry, I'm just a stats junkie

-----

Here is one of my favorite 'riddles' :

Say that a certain game show has a feature where a contestant is presented with three doors. A worthwhile prize is behind only one of the doors. The contestant is instructed to pick one of the three doors. Afterwards, the host always opens one of the remaining doors that does not have the prize behind it, and then questions the contestant about whether they want to remain with their chosen door or switch to the remaining door.

The question - What is the rule of thumb that provides a higher 'statistical' chance of winning?

sorry, I'm just a stats junkie

Posted: **Wed Oct 27, 2004 11:31 am**

In germany the show was/is called "Geh auf's Ganze".

At the moment, i'm too lazy to do the math, but the contestant is wise if he always change his mind. He should always chose the door, which the host left in the game.

But the the german show, those steps were changed Now, the contestant has also to chose which of the left doors, the host should open. this way the likelyhood is always the same.

At the moment, i'm too lazy to do the math, but the contestant is wise if he always change his mind. He should always chose the door, which the host left in the game.

But the the german show, those steps were changed Now, the contestant has also to chose which of the left doors, the host should open. this way the likelyhood is always the same.

Posted: **Wed Oct 27, 2004 11:36 am**

He should switch doors. I forget the specifics, but it comes down to the probability given the host has selected one of the remaining two doors, making the other one more likely to contain a prize than the original door picked by the contestant.Pink wrote:The question - What is the rule of thumb that provides a higher 'statistical' chance of winning?

Posted: **Wed Oct 27, 2004 2:22 pm**

When he did the first choice, he had a 1/3 chance of picking the right door. The secound time one of the false doors are eliminated so he has 1/2 chance of picking the right door if he changes. It is either the door he took from the beginning or the other one. But it's still only 1 in 3 that the first door he choosed has the prize.

Posted: **Wed Oct 27, 2004 3:08 pm**

@Zpock:

You haven't seen the show, have you?

there are two prices and one "zonk" So if you change your mind, you will ALWAYS get an price, not necessarly the main one.

You haven't seen the show, have you?

there are two prices and one "zonk" So if you change your mind, you will ALWAYS get an price, not necessarly the main one.

Posted: **Wed Oct 27, 2004 11:10 pm**

Didn't know that it was an actual show. Then again, my choices of German stations over here is a bit limited

Posted: **Thu Oct 28, 2004 3:50 pm**

Well, as with most of the german shows, the original is a show from the US. Forgot the name, though. And in this show there were 2 Goats and 1 Price...

And here is a different proof that the chance of winning is 2/3 if you switch doors:

Let's assume that the winning door is A, then depending on the door you chose first, the way to win is:

If you chose A, stay.

If you chose B, switch.

If you chose C, switch.

(keep in mind that after the moderator opened a door with a goat, there are only 2 left)

So, in 2 of 3 cases, switching is better.

And if you still don't believe it, think of it in this way: Just assume that you have 1000 doors instead of 3. Only 1 win, 999 blanks. After you chose one door, the hosts openes 998 doors with blanks.

Do you still believe that you got lucky in the first guess and have chosen correctly (chance 1:1000), or do you think the remaining door holds the price?

And here is a different proof that the chance of winning is 2/3 if you switch doors:

Let's assume that the winning door is A, then depending on the door you chose first, the way to win is:

If you chose A, stay.

If you chose B, switch.

If you chose C, switch.

(keep in mind that after the moderator opened a door with a goat, there are only 2 left)

So, in 2 of 3 cases, switching is better.

And if you still don't believe it, think of it in this way: Just assume that you have 1000 doors instead of 3. Only 1 win, 999 blanks. After you chose one door, the hosts openes 998 doors with blanks.

Do you still believe that you got lucky in the first guess and have chosen correctly (chance 1:1000), or do you think the remaining door holds the price?

Posted: **Thu Oct 28, 2004 5:30 pm**

Another alalogy, a bag with 2 Black and 1 white marble. Reach in and take out a marble and keep it in your fist without looking at it. A Blck Marble is then removed from the bag, do you switch the one in your hand with the one remaining in the bag to get the white marble?

Posted: **Wed Nov 03, 2004 12:11 pm**

Another one:

You have two metal rods, one is magnetic, the other is not. How do you find out which is which? (no additional tools allowed)

You have two metal rods, one is magnetic, the other is not. How do you find out which is which? (no additional tools allowed)

Posted: **Wed Nov 03, 2004 12:50 pm**

the magnetic one isn't magnetic at the middel, if you cross both rods on the middel, they don't hold together. if you move one rod by keeping the other one at the middel you can discover which one is magnetic.

terrible english

terrible english